\newproblem{lay:4_5_1}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.5.1}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Find a basis for the subspace below and state its dimension
	\begin{center}
		$S=\left\{\begin{pmatrix}s-2t\\s+t\\3t\end{pmatrix}\quad \forall s,t\in\mathbb{R}\right\}$
	\end{center}
}{
  % Solution
	We may write the set as
	\begin{center}
		$S=\left\{s\begin{pmatrix}1\\1\\0\end{pmatrix}+t\begin{pmatrix}-2\\1\\3\end{pmatrix}\quad \forall s,t\in\mathbb{R}\right\}$\\
	\end{center}
	Thus, a basis is given by the vectors
	\begin{center}
		$\mathrm{Basis}\{S\}=\left\{\begin{pmatrix}1\\1\\0\end{pmatrix},\begin{pmatrix}-2\\1\\3\end{pmatrix}\right\}$ \\
	\end{center}
	Since the basis has two vectors, the dimension of $S$ is 2.
}
\useproblem{lay:4_5_1}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
